Two circles C1 and C2 intersect at points A and B. Let P, Q be points on circles C1, C2 respectively, such that |AP|=|AQ|. The segment PQ intersects circles C1 and C2 in points M, N respectively. Let C be the center of the arc BP of C1 which does not contain point A and let D be the center of arc BQ of C2 which does not contain point A Let E be the intersection of CM and DN. Prove that AE is perpendicular to CD. Proposed by Steve Dinh
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Tags: geometry, emc
01.01.2017 07:17
[asy][asy]size(7.5cm);pair A=(3.46,4.04), B=(2.79,-0.35), P=(0.39,6.05), Q=(4.73,0.59), M=(3.84,1.71), C=(-3.40,0.98), D=(3.88,-0.13),Nn=(2.06,3.95); pair Ee=extension(C,M,D,Nn); draw(circumcircle(A,B,C)^^circumcircle(A,B,D),green); D(MP("A",A,NE)--MP("P",P,N)--MP("N",Nn,NE)--MP("M",M,E)--MP("Q",Q,SE)--A,heavymagenta); D(A--MP("C",C,SW)--MP("D",D,S)--A--MP("B",B,S),heavycyan); draw(D--Nn^^C--M,darkmagenta+dashed); dot(A);dot(B);dot(C);dot(D);dot(P);dot(Q);dot(M);dot(Nn);D(MP("E",Ee,SW),black); [/asy][/asy] Okay, let's chase down some angles. ∠CAD=∠CAB+∠BAD=12∠PAB+12∠BAQ=12∠PAQ=90∘−∠AQN (since △APQ is A−isosceles)=90∘−∠ADN⟹∠CAD=90∘−∠ADNThis implies DN⊥AC; similarly CM⊥AD, so E is in fact the orthocenter of △ACD. The fact that AE⊥CD follows easily. ◼
25.12.2021 17:50
Easy one we will prove E is orthocenter of ACD. ∠PAC = ∠CAB and ∠BAD = ∠DAQ and PAQ is isosceles so ∠CAD + ∠AQP = 90. ∠CAD + ∠AQP = ∠CAD + ∠ADN = 90 ---> DE ⊥ AC. ∠CAD + ∠APQ = ∠CAD + ∠ACM = 90 ---> CE ⊥ AD. so E is orthocenter and we're Done.